Mathcad - Next100 PV 316Ti Incshuman/NEXT/CURRENT_DESIGN/PV/calc_Next10… · Design Rules: ASME...
Transcript of Mathcad - Next100 PV 316Ti Incshuman/NEXT/CURRENT_DESIGN/PV/calc_Next10… · Design Rules: ASME...
MAWPpv 15.4 bar=
P MAWPpv:=
Mass supported internally by pressure vessel
Internal copper shield (ICS)
MICS_cyl 6000kg:= MICS_eh 1500kg:= MICS_tp 2500kg:=
Detector subsystems, est.
Mep 750kg:= Mtp 200kg:= Mfc 350kg:=
Length inside vessel of copper, total
LCu 2.0m:=
Mass total of internal copper shielding:
MICS MICS_cyl MICS_eh+ MICS_tp+:= MICS 10000 kg=
Maximum mass supported on internal flange of each head:
Mfl_h MICS_tp:=
Maximum mass supported on each internal flange of the main cylindrical vessel:
Mfl_v 0.5 MICS_cyl Mfc+( ) Mep+:= Mfl_v 3925 kg= this mass will be present when heads are not mounted
Estimated approximate total vessel mass carried on supports (numbers from calcs below):
Active volume dimensions, from earlier analyses:
rXe 0.53 m= lXe 1.3 m=
We consider using a field cage solid insulator/light tube of 3 cm total thk., and a copper liner of 12 cm thickness,
including all tolerances and necessary gaps.
tfc 3cm:= tCu 12cm:=
Pressure Vessel Inner Radius, Diameter:
Ri_pv rXe tfc+ tCu+:= Ri_pv 0.68 m= Di_pv 2Ri_pv:= Di_pv 1.36 m=
Pressure vessel length:
main cyl. vessel overall, inside
Lv 1.6m:= Lo 2.2m:=
Temperatures:
For pressure operation, the temperature range will be 10C-30C. For vacuum operation, the temperature range will
be 10C to 150C (bakeout).
Maximum Operating Pressure (MOP), gauge:
MOPpv PMOPa 1bar−( ):= MOPpv 14 bar=
Minimum Pressure, gauge:
Pmin 1.5− bar= the extra 0.5 atm maintains an upgrade path to a water or scintillator tank
Maximum allowable pressure, gauge (from LBNL Pressure Safety Manual, PUB3000)
From LBNL PUB3000, recommended minimum is, 10% over max operating pressure; this is design pressure at
LBNL. This is for spring operated relief valves, to avoid leakage. Use of pilot operated relief valves can reduce this to
as little as 2%, as they seal tighter when approaching relief pressure:
MAWPpv 1.1MOPpv:=
vessel heads flanges ρSS 8
gm
cm3
:=
Mv ρSS 2πRi_pv 10⋅ mm Lo⋅ πRi_pv2
12⋅ mm+ 4 2⋅ πRi_pv 4.2⋅ cm 5⋅ cm+
⋅:= Mv 1179 kg=
Total detector mass:
Mdet MICS Mep+ Mfc+ Mtp+ Mv+:= Mdet 1.248 104
× kg=
Vessel wall thickness, for internal pressure is then (div. 1), Assume all welds are type (1) as defined in
UW-12 , are double welds, fully radiographed, so weld efficiency:
E 1:=
Minimum wall thickness is then:
tpv_d1_min_ip
P Ri_pv⋅
S E⋅ 0.6 P⋅−:=
tpv_d1_min_ip 7.75 mm=
We set wall thickness to be:
tpv 10mm:= tpv tpv_d1_min_ip> 1=
Maximum Allowable External Pressure
ASME PV code Sec. VIII, Div. 1- UG-28 Thickness of Shells under External Pressure
Maximum length between flanges Lff 1.6m:=
The maximum allowable working external pressure is determined by the following procedure:
Compute the following two dimensionless constants:
Lff
2Ri_pv
1.2=2Ri_pv
tpv
136=
From the above two quantities, we find, from fig. G in subpart 3 of Section II, the factor A:
Vessel wall thicknesses
Material:
We use 316Ti for vessel shells and flanges due to its known good radiopurity and strength.
Design Rules:
ASME Boiler and Pressure Vessel code section VIII, Rules for construction of Pressure vessels
division 1 (2010)
316Ti is not an allowed material under section VIII, division 2, so we must use division 1 rules. The saddle
supports are however, deisgned using the methodology given in div. 2, as div. 1 does not provide design formulas
(nonmandatory Appendix G)
Maximum allowable material stress, for sec. VIII, division 1 rules from ASME 2009 Pressure Vessel
code, sec. II part D, table 1A:Youngs modulus
Smax_316Ti_div1 20000psi:= -20F - 100F ESS_aus 193GPa:=
color scheme for this document
input check result (all conditions should be true (=1)
xx 1:= xx 0> 1=
Choose material, then maximum allowable strength is:
S Smax_316Ti_div1:=
A. 0.0005:=
Using the factor A in chart (HA-2) in Subpart 3 of Section II, Part D, we find the factor B ( @
400F, since we may bake while pulling vacuum):
B. 6200psi:= @ 400F
The maximum allowable working external pressure is then given by :
Pa
4B.
32Ri_pv
tpv
:=Pa 4.1 bar= Pmin− 1.5 bar=
Pa Pmin−> 1=
From sec. VIII div 1, non-mandatory appendix Y for bolted joints having metal-to-metal contact outside of
bolt circle. First define, per Y-3:
Sb 255.1MPa=Sb Smax_N07718:=
Smax_N07718 37000psi:=Inconel 718 (UNS N07718)
Maximum allowable design stress for bolts, from ASME 2010 Pressure Vessel code, sec. II part D, table 3
Sf 2 104
× psi=Sf 137.9 MPa=Sf Smax_316Ti_div1:=
Maximum allowable design stress for flange
Maximum allowable material stresses, for sec VIII, division 1 rules from ASME 2010 Pressure Vessel
code, sec. II part D, table 2A (division 1 only):
The flange design for O-ring sealing (or other self energizing gasket such as helicoflex) is "flat-faced", with
"metal to metal contact outside the bolt circle". This design avoids the high flange bending stresses found
in a raised face flange (of Appendix 2) and will result in less flange thickness. The rules for this design are
found only in sec VIII division 1 under Appendix Y, and must be used with the allowable stresses of
division 1. Flanges and shells will be fabricated from 316Ti (ASME spec SA-240) stainless steel plate.
Plate samples will be helium leak checked before fabrication, as well as ultrasound inspected for flat
laminar flaws which may create leak paths. The flange bolts and nuts will be inconel 718, (UNS N77180)
as this is the highest strength non-corrosive material allowed for bolting.
We will design with enough flange strength to accomodate using a Helicoflex 5mm gasket
(smallest size possible) specially designed with a maximum sealing force of 70 N/mm.
(gauge pressure)P 15.4 bar=Ri_pv 0.68 m=
max. allowable pressureinner radius
Flange thickness, head to vessel main flanges:
68.65 2⋅ 137.3=O-ring mean radius as measured in CAD model:G 1.373m=G 2 Ri_pv .65cm+( ):=
Gasket dia
C 1.43 m⋅:=
Bolt circle (B.C.) dia, C:
B1 1.37 m=B1 B g1+:=
define:
B 1.36 m=B 2Ri_pv:=
Flange ID
A 1.48m:=
Flange OD
r1 5 mm=r1 max .25g1 5mm,( ):=g1 10 mm=g0 10 mm=g1 tpv:=g0 tpv:=
corner radius:hub thickness at flange (no hub)
using nomenclature and formulas from this chart at http://www.tribology-abc.com/calculators/metric-iso.htm
h3 .6134 pt⋅:=pt 1.0mm:=
Choosing ISO fine thread, to maximize root dia.; thread depth is:
nmax 140=nmax truncπC
2.0db
:=maximum number of bolts possible,
using narrow washers:db 16mm:=n 132:=
Start by making trial assumption for number of bolts, nominal bolt dia., pitch, and bolt hole dia D,
Helicoflex recommends using Y2 ( 220 N/mm) for large diameter seals, even though for small diameter one
can use the greater of Y1 or Ym=(Y2*(P/Pu)). For 15 bar Y1 is greater than Ym but far smaller than Y2.
Sealing is less assured, but will be used in elastic range and so may be reusable. Flange thickness and bolt
load increase quite substantially when using Y2 as design basis, which is a large penalty. We plan to
recover any Xe leakage, as we have a second O-ring outside the first and a sniff port in between, so we thus
design for Y1 (use Fm) and "cross our fingers" : if it doesn't seal we use an O-ring instead and recover
permeated Xe with a cold trap. Note: in the cold trap one will get water and N2, O2, that permeates through
the outer O-ring as well.
Fj 9.489 105
× N=Fm 3.019 105
× N=
or Fj π Dj⋅ Y2⋅:=Fm πDj Y1⋅:=
Force is then either of:
Dj 1.373 m=Dj G:=for gasket diameter
Y2 220N
mm:=Y1 70
N
mm:=
recommended value for large diameter seals,
regardless of pressure or leak rate
min value for our pressure
and required leak rate (He)
for 5mm HN200 with aluminum jacket:
for O-ring, 0.275" dia., shore A 70 F= ~5 lbs/in for 20% compression, (Parker O-ring handbook); add
50% for smaller second O-ring. (Helicoflex gasket requires high compression, may damage soft Ti
surfaces, may move under pressure unless tightly backed, not recommended)
Helicoflex has equivalent formulas using Y as the unit force term and gives several possible values.
Sealing force, per unit length of circumference:
H 2.31 106
× N=H .785G2
MAWPpv⋅:=
Force of Pressure on head
Note: this diameter will be correct for Helicoflex gasket, but slightly higher for O-ring, which is fluid and
"transmits pressure" out to its OD, howqever the lower gasket unit force of O-ring more than compensates, as
per below:
We will thread some of these clearance holes for M20-1.5 bolts to allow the head retraction fixture to be bolted
up the the flange. The effective diameter of these holes will be the average of nominal and minimum diameters.
To avoid thread interference with flange bolts, the studs will be machined to root diameter per UG-12(b).in
between threaded ends of 1.5x diameter in length. The actual clearance holes will be 18mm, depending on
achievable tolerances, so as to allow threading where needed.
Dtmin 18 mm=Dtmin db 2mm+:=
Flange hole diameter, minimum for clearance :
0.5C 0.5B g1+ r1+( )− 0.5ODw≥ 1=
this is for standard narrow washers, and for wrench
sockets which more than cover the nut width across
corners
ODw 2db:=Check nut, washer, socket clearance:
πC
n34.034 mm=actual bolt to bolt distance:πC 2.0n db⋅− 0≥ 1=
Check bolt to bolt clearance, here we use narrow thick washers (28mm OD) under the 24mm wide (flat to flat)
nuts (28mm is also corner to corner distance on nut), we adopt a minimum bolt spacing of 2x the nominal bolt
diameter (to give room for a 24mm socket) :
Ab 226.263 cm2
=Ab nπ
4⋅ d3
2:=
Total bolt cross sectional area:
d3 14.7732 mm=d3 db 2h3−:=
Bolt root dia. is then:
<--- use H1 for 1.5mm pitch
<---use h3 for 1.0 mm pitch
HT 4.353 104
× N=
hT 0.5 R. g1+ hG+( )⋅:= hT 31.75 mm= from Table 2-6 Appendix 2, int. fl.
Total Moment on Flange
MP HD hD⋅ HT hT⋅+ HG hG⋅+:= MP 7.797 104
× J=
Appendix Y Calculation
P 15.4 bar=
Choose values for plate thickness and bolt hole dia:
t 4.15cm:= D Dt:= D 1.919 cm=
Going back to main analysis, compute the following quantities:
βC B1+
2B1
:= β 1.022= hC 0.5 A C−( ):= hC 2.5 cm=
ARn D⋅
π C⋅:= AR 0.564= h0 B g0⋅:= h0 11.662 cm=
aA C+
2B1
:= a 1.062=
rB1
n
4
1 AR2
−
atan1 AR+
1 AR−
π− 2AR−
:= rB 7.462 103−
×=
We need factors F and V, most easily found in figs 2-7.2 and 7.3 (Appendix 2)
H1 .812mm:= from chart above
dmin_20_1.5 20mm 2 H1⋅−:= dmin_20_1.5 1.838cm= this will be max bolt hole size or least material
condition (LMC)
dmin_20_1.5 Dtmin≥ 1=
De 0.5 20 mm⋅ dmin_20_1.5+( ):= De 1.919 cm=
Set:
Dt De:= Dt Dtmin> 1=
Compute Forces on flange:
We use a unit gasket seating force of Y1 above
HG Fm:= HG 3.019 105
× N=
hG 0.5 C G−( ):= hG 2.85 cm= from Table 2-6 Appendix 2, Integral flanges
HD .785 B2
⋅ P⋅:= HD 2.266 106
× N=
R. 0.5 C B−( ) g1−:= R. 2.5 cm= radial distance, B.C. to hub-flange intersection, int fl..
hD R. 0.5g1+:= hD 3 cm= from Table 2-6 Appendix 2, Int. fl.
HT H HD−:=
E ESS_aus:= E 193 GPa= EInconel_718 208GPa:= Ebolt EInconel_718:=
Flange Moment due to Flange-hub interaction
(7)MS
JP− F'⋅ MP⋅
t3
JS F'⋅+
:= MS 1.8− 103
× N m⋅=
Slope of Flange at I.D.
opening half gap =θB
5.46
E π⋅ t3
JS MS⋅ JP MP⋅+( ):= θB 5.903 104−
×= (8)θB 3⋅ cm 0.018 mm=
E θB⋅ 113.924 MPa=Contact Force between flanges, at hC:
(9)HC
MP MS+
hC
:= HC 3.045 106
× N=
Bolt Load at operating condition:
Wm1 H HG+ HC+:= Wm1 5.657 106
× N= (10)
Operating Bolt Stress
σb
Wm1
Ab
:= σb 250 MPa= Sb 255.1MPa= (11)
rEE
Ebolt
:= rE 0.928= elasticity factor
since g1
g0
1= these values converge to F 0.90892:= V 0.550103:=
Y-5 Classification and Categorization
We have identical (class 1 assembly) integral ( category 1) flanges, so from table Y-6.1, our applicable
equations are (5a), (7) - (13), (14a), (15a), (16a)
JS1
B1
2 hD⋅
β
hC
a+
πrB+:= JS 0.083= JP1
B1
hD
β
hC
a+
π rB⋅+:= JP 0.062=
(5a)MP 7.797 10
4× N m⋅=
F'g0
2h0 F t⋅+( )
V:= F' 2.806 10
5−× m
3=
A 1.48 m= B 1.36 m=
KA
B:= K 1.088= Z
K2
1+
K2
1−
:= Z 11.854=
f 1:= hub stress correction factor for integral flanges, use f =1 for g1/g0=1 (fig 2-7.6)
ts 0mm:= no spacer between flanges
l 2t ts+ 0.5db+:= l 9.1 cm= strain length of bolt ( for class 1 assembly)
Y-6.1, Class 1 Assembly Analysis
http://www.hightempmetals.com/techdata/hitempInconel718data.php
Elastic constants:
σb Sb< 1=
(b)
(1) SH 1.5Sf< 1= Snnot applicable
(2) not applicable
(c) SR_BC Sf< 1=
SR_ID Sf< 1=
(d) ST Sf< 1=
(e) SH SR_BC+
2Sf< 1=
SH SR_ID+
2Sf< 1=
(f) not applicable
Bolt force
Fbolt σb .785⋅ db2
⋅:= Fbolt 5.024 104
× N=
Bolt torque required, minimum:
Design Prestress in bolts
Si σb
1.159 hC2
⋅ MP MS+( )⋅
a t3
⋅ l rE⋅ B1⋅
−:= Si 243.7 MPa= (12)
Radial Flange stress at bolt circle
SR_BC
6 MP MS+( )
t2
π C⋅ n D⋅−( ):= SR_BC 135.4 MPa= (13)
Radial Flange stress at inside diameter
SR_ID
2F t⋅
h0 F t⋅+6+
−MS
πB1 t2
⋅
⋅:= SR_ID 1.61 MPa= (14a)
Tangential Flange stress at inside diameter
(15a)ST
t E⋅ θB⋅
B1
2F t⋅ Z⋅
h0 F t⋅+1.8−
MS
πB1 t2
⋅
⋅+:= ST 2.46 MPa=
Longitudinal hub stress
(16a)SH
h0 E⋅ θB⋅ f⋅
0.91g1
g0
2
B1 V⋅
:= SH 19.372 MPa=
Y-7 Bolt and Flange stress allowables: Sb 255.1MPa= Sf 137.9 MPa=
(a)
Additional Calculations for Shielding Weight:
Shear stress in inner flange lip from shield (could happen only if flange bolts come loose, are left loose, or if joint
opens under pressure, otherwise friction of faces will support shield, given additional tension, as permissible under
non-mandatory Appendix S above )
Masses of Copper shielding in cyl and heads (maybe extra in tracking head)
tCu 0.12 m= tCu_h 20cm:= Lff 1.6 m= ρCu 9 103
×kg
m3
=
Msh_head ρCu π⋅ Ri_pv2
tCu_h⋅:= Msh_head 2.615 103
× kg=
Msh_cyl ρCu 2⋅ π Ri_pv⋅ tCu⋅ Lff⋅:= Msh_cyl 7.383 103
× kg=
Msh Msh_cyl 2Msh_head+:= Msh 1.261 104
× kg= slightly less than this, due to gaps
tlip 3mm:=
Shear stress in lip (projected force):
τ lip
Msh_head g⋅
Ri_pv tlip⋅:= τ lip 12.57 MPa=
Shear stress on O-ring land (section between inner and outer O-ring), from pressurized O-ring. This is assumed to be
the primary stress. There is some edge moment but the "beam" is a very short one. This shear stress is not in the
same direction as the nominal tangential (hoop) stress of the flange.
for pressure test use 1.5x
this valueTbolt_min 0.2Fbolt db⋅:= Tbolt_min 160.8 N m⋅= Tbolt_min 118.6 lbf ft⋅=
This is the minimum amount of bolt preload needed to assure joint does not open under pressure. An additional
amount of bolt preload is needed to maintain a minimum frictional shear resistance to assure head does not slide
downward from weight; we do not want to depend on lip to carry this. Non-mandatory Appendix S of div. 1 makes
permissible higher bolt stresses than indicated above when needed to assure full gasket sealing and other
conditions. This is consistent with proper preloaded joint practice, for properly designed joints where connection
stiffness is much greater than bolt stiffness, and we are a long way from the yield stress of the bolts
Mhead 2500kg:= µSS_SS .7:= typ. coefficient of friction, stainless steel (both) clean and dry
Vhead Mhead g⋅:= Vhead 2.452 104
× N=
Fn
Vhead
µSS_SS
:= Fn 3.502 104
× N= this is total required force, force required per bolt is:
Fn_bolt
Fn
n:= Fn_bolt 265.331 N= this is insignificant compared to that required for pressure.
Let bolt torque for normal operation be then 25% greater than minimum:
Tbolt 1.25Tbolt_min:= Tbolt 201 N m⋅= Tbolt 148 ft lbf⋅=
It is recommended that a pneumatic torque wrench be used for tightening of bolts. Anti-seize lubricant (checked for
radiopurity) should be used on threads and washers. Bolts should be tightened in 1/3 full torque increments, but
there is no specific tightening pattern to be used, as gasket compression is not determined by bolt tightness. The
head lift fixture may be retracted once all bolts not occupied by lift fixture have been tightened to the first 1/3
torque increment; there will be adequate frictional shear resistance to eliminate head slippage while detaching lift
fixture. Bolts should be run up uniformly to fully close gap before proceeding with tightening. Do not forget to install
sleeves in all threaded holes after removing lift fixture.
Ssy_65500_H2 131 MPa=Ssy_65500_H2 0.5Sy_65500_H2:=
Sy_65500_H2 262.001 MPa=Sy_65500_H2 38000psi:=
This stress is inconsequential, as bolts will be ASME SB-98 silicon copper UNS C65500 - HO2 (half hard cond);
this material should be radiopure and has > 20% elongation in the hard condition. Shear strength in yield is 50%
Sy.
τbolt_cubar 12.347 MPa=τbolt_cubar
0.5Mcubar_vfan g⋅
5π
4⋅ droot_M6
2:=
Mcubar_vfan 225kg:=
On the tracking side, the bars will be pulled up tight to the inside flange. On the energy side
they must float axially, this is done using a special shoulder bolt which provides a loose
double shear connection. Worst case would be single shear, where the tracking side bolts
are left loose.
droot_M6 4.77mm:=
The internal copper shield bars are attached to the inside flanges with M6-1 bolts. The worst case for attachment
ar the bars with collimation holes; these are narrow where they attach. For a flange hole pattern of 240 bolts, there
are 5 attachment holes at each end.
Bolt loads from Cu bars
τ land 1.778MPa=τ land
FO_ring_land
AO_ring_land
:=
AO_ring_land 2πRi_pv tland_radial⋅:=
FO_ring_land 2πRi_pv wland_axial⋅ P⋅:=
wland_axial .41cm:=tland_radial .36cm:=
Strain, O-ring cross section, in axial direction
εOpa 0.5− εOpt:= εOpa 0.016−=
O-ring dia., stretched:
dOps dOp 1 εOpa+( )⋅:= dOps
4.918
5.253
mm=
Resulting squeeze (using the vectorize operator to continue parallel calculations)
sqp
dOps dOpg−
dOps
→
:= sqp
22.74
27.659
%= 15% sqp< 30%<1
1
=
O-ring groove cross sectional area,
AOpg dOpg ROgpo ROgpi−( )⋅1
2
π
2−
rip2
⋅−
→
:= AOpg 2.558 105−
× m2
=
Trelleborg recommends no more than 85% fill ratio
Rfp
π
4dOps
2
AOpg
→
:= Rfp
74.274
84.718
%= Rfp 85%<1
1
=
Outer (vacuum) O-ring:
Groove wall radii (average), depth, inner corner radii:
O-Ring groove dimensions
the Recommmended range of compression for static face seals is 21-30% in the Parker O-ring handbook;
Trelleborg recommend 15-30%. For each nominal size, there are several cross sections, metric, JIS and
A-568. Ity ios recommended by this author to design a groove which can accomodate all these cross sections
with squezze in the acceptable range, so as to give the most flexibility.
For large diameter O-rings, Parker recommends using one size smaller to avoid sag. This is feasible
for the inner O-ring, as the undercut lip is on the ID of the groove, but will not work on the outer vacuum O-ring
as the undecut must be on the OD (otherwise the undercut may reduce seal effectiveness). Using an O-ring 1
or 2 sizes larger on the outer O-ring may develop enough compressive stress to retain O-ring in groove, but this
should be tested. Stiffer compounds may help here if there is a problem Regardless, the groove dimensions
should account for the stretch or compression of the O-ring which changes its effective cross section diameter.
There are several close sizes that Trelleborg makes unspliced O-rings from (these are strongly preferred) and a
stiffer than normal compound could be used for the vacuum O-ring, if needed
Inner (pressure bearing) O-ring:
Groove wall radii (average), depth, inner corner radii:
ROgpo 688.7mm:= ROgpi 682.25mm:= dOpg 3.8mm:= rip 1mm:=
O-ring inner radius, cross section diameter, unstretched
metric sizeROpi 660mm:= dOp
5
5.34
mm:=AS - 568 size
O-ring elongation (tangential direction, normal to cross section)
εOpt 1ROpi
ROgpi
−:= εOpt 3.261 %= recommeded less than 3% (Trelleborg); 3% is our min. target
Bulk Modulus of most rubber polymers is very high, material is essentially incompressible (Poisson's ratio = -0.5)
εOva 0.027=
O-ring dia., stretched:
dOvs dOv 1 εOva+( )⋅:= dOvs
3.08
3.645
mm=
Resulting squeeze
sqv
dOvs dOvg−
dOvs
→
:= sqv
15.591
28.669
%= 15% sqv< 30%<1
1
=
O-ring groove cross sectional area,
AOvg dOvg ROgvo ROgvi−( )⋅1
2
π
2−
riv2
⋅−
→
:= AOvg 1.268 105−
× m2
=
Fill ratio; Trelleborg recommends no more than 85%:
We should have a comfortable margin hereRfv
π
4dOvs
2
AOvg
→
:= Rfv
58.752
82.269
%= Rfv 85%<1
1
=
ROgvo 697.66mm:= ROgvi 692.93mm:= dOvg 2.6mm:= riv 0.6mm:=
O-ring inner radius, cross section diameter, unstretched
metric size note: there are several intermediate sizes ROvi 730mm:= dOv
3
3.55
mm:=metric/JIS size
O-ring elongation (tangential direction, normal to cross section)
εOvt 1ROvi
ROgvi
−:= εOvt 5.35− %= recommended less than 3% (Trelleborg); we go for ~5% here as
compression should not compromise integrity
Bulk Modulus of most rubber polymers is very high, material is essentially incompressible (Poisson's ratio = -0.5)
Strain, O-ring cross section, in axial direction
εOva 0.5− εOvt:=
Support Design using rules of div 2, part 4.15:
From the diagram below the rules are only applicable to flange attached heads if there is a flat cover or
tubesheet inside, effectively maintaining the flanges circular. Since the PMT carrier plate and shielding is firmly
bolted in, it serves this purpose and we may proceed. We must also compute the case with the heads
attached, as there will be additional load
a) Design Method- although not specifically stated, the formulas for bending moments at the center and at the
supports are likely based on a uniform loading of the vessel wall from the vessel contents. In this design, the internal
weight (primarily of the copper shield) is applied at the flanges; there is no contact with the vessel shell. We
calculate both ways and take the worst case.
M1' 1.706 104
× N m⋅=
M2' M1':= M2' 1.706 104
× N m⋅=
TQ L 2a−( )⋅
L4h2
3+
:= T 3.215 104
× N=
4.15.3.3 - longitudinal stresses
distributed load (ASME assumption) end load ( actual)
σ1
P Rm⋅
2tpv
M2
π Rm2tpv
−:= σ1 52.789 MPa= σ1'
P Rm⋅
2tpv
M2'
π Rm2tpv
−:= σ1' 52.301 MPa=
σ2'
P Rm⋅
2tpv
M2'
π Rm2tpv
+:= σ2' 54.616 MPa=σ2
P Rm⋅
2tpv
M2
π Rm2tpv
+:= σ2 54.128 MPa=
same stress at supports, since these are stiffened, as a<0.5Rm and close to a torispheric head a 0.5Rm< 1=
σ3
P Rm⋅
2tpv
M1
π Rm2tpv
−:=σ3 53.345 MPa= σ3'
P Rm⋅
2tpv
M1'
π Rm2tpv
−:= σ3' 52.301 MPa=
L Lff:= Mtot 12000kg:= L 1.6 m=
b 1.5cm:= amin .18Lff:= amin 28.8 cm= a 29cm:= θ 120deg:= Rm Ri_pv 0.5tpv+:=
b1 min b 1.56 Rm tpv⋅⋅+( ) 2 a⋅, := b1 14.411 cm= k 0.1:=h2 20cm:=
θ1 θθ
12+:= θ1 130 deg= maximum reaction load at each support:
Q 0.5Mtot g⋅:= Q 5.884 104
× N=
M1 Q− a⋅ 1
1a
L−
Rm2
h22
−
2 a⋅ L⋅+
14h2
3L+
−
⋅:= M1 1.676 103
× N m⋅= Q a⋅ 1.706 104
× J=
M2
Q L⋅
4
12 Rm
2h2
2−
⋅
L2
+
14 h2⋅
3L+
4a
L−
⋅:= M2 9.875 103
× N m⋅=
M1' Q a⋅:=
K5 0.76=K51 cos α( )+
π α− sin α( ) cos α( )⋅+:=
4.15.3.5 Circumferential Stress
(4.15.14)τ1 1.749MPa=τ1
K2 T⋅
πRm tpv⋅:=c)
here we use c), formula for cyl. shell with no stiffening rings and which is not stiffened by a formed head, flat
cover or tubesheet. This is worst case, as we havea flange, which can be considered as one half of a
stiffening ring pair for each support.
K2 1.171=K2sin α( )
π α− sin α( ) cos α( )+:=
α 1.99=α 0.95 πθ
2−
:=
∆ 1.396=∆π
6
5θ
12+:=
4.15.3.4 - Shear stresses
σ4' 54.616 MPa=σ4'
P Rm⋅
2tpv
M1'
π Rm2tpv
+:=σ4 53.572 MPa=σ4
P Rm⋅
2tpv
M1
π Rm2tpv
+:=
b1 14.411 cm=
σ7Q−
4tpv b x1+ x2+( )⋅
12K7 Q⋅ Rm⋅
L tpv2
⋅
−:= σ7 156.484 MPa= (4.15.25)
too high; we need a reinforcement plate of thickness;
tr tpv:= strength ratio: η 1:= (4.15.29)
σ7rQ−
4 tpv η tr⋅+( ) b1⋅
12K7 Q⋅ Rm⋅
L tpv η tr⋅+( )2
⋅
−:= σ7r 36.569 MPa= (4.15.28)
3)f) Acceptance Criteria
S 1.379 108
× Pa= S 2 104
× psi=
σ7r 1.25S< 1=
4) this section not applicable as tr 2tpv> 0=
β πθ
2−:= β 2.094=
K6
3 cos β( )⋅
4
sin β( )β
2
⋅5 sin β( )⋅ cos β( )⋅
4 β⋅−
cos β( )3
2+
sin β( )4 β⋅
−cos β( )
4+ β sin β( )⋅
sin β( )β
2
1
2−
sin 2 β⋅( )4 β⋅
−
⋅−
2 π⋅sin β( )
β
2
1
2−
sin 2 β⋅( )4 β⋅
−
⋅
:=
K6 0.221−=
a
Rm
0.5< 1=
K7
K6
4:= K7 0.055−=
a) Max circ bending moment
1) Cyl shell without a stiffening ring
Mβ K7 Q⋅ Rm⋅:= Mβ 2.223− 103
× N m⋅=
c) Circ. stress in shell, without stiffening rings
x1 0.78 Rm tpv⋅:= x1 6.456 cm= x2 x1:= k 0.1=
σ6
K5− Q⋅ k⋅
tpv b x1+ x2+( )⋅:= σ6 3.104− MPa=
L 8Rm< 1=L 1.6 m=
4.15.3.6 - Saddle support, horizontal force given below must be resisted by low point of saddle ( where height = hs)
Fh Q1 cos β( )+ 0.5 sin β( )
2⋅−
π β− β sin β( )⋅ cos β( )+
⋅:= Fh 5.242 104
× N= hs 9cm:=
σh
Fh
b hs⋅:= σh 38.833 MPa=
Support on, and Attachment to floor
The vessel has four points of connection ("corners") to the seismic platform, two on each saddle support; each
connection point consisting of two M16mm studs. Other possibilities exist. Pressurization or thermal excursion
(bakeout, cryogen spill from ArDM) will result in dimensional changes of the vessel, so it is required to use low
friction pads under the supports to constrain the vessel in a 2D kinematic fashion. One corner is fixed, two others
are slotted to allow sliding in one direction (orthogonal to each other), and a full clearance hole pattern at the fourth
corner allows sliding in both directions.
σhoop 106.137 MPa=σhoop
P Ri_pv⋅
tpv
:=σ long 53.041 MPa=σ long
HD
2πRi_pv tpv⋅:=
stresses in vessel shell, longitudinal and tangential (hoop):
ws 1.2m:=
saddle support width, transverse
HD 2.266 106
× N=tpv 10 mm=Ri_pv 0.68 m=Ls 1.02 m=Ls Lff 2a−:=
length between saddle supports:Pressure load, longitudinal
Vessel length and width change under pressurization and heating:
These deflections (from either pressure or thermal excursion) are substantial enough to warrant the use of low
friction pads under three of the four supports, which will allow the vessel to slide both lengthwise and widthwise
when pressurizing/depressurizing or baking. In addition there is a remote possibility of cryogen spillage, perhaps
from ArDM which may chill the vessel, so a capacity for contraction equal to the above expansion should be
designed in. Bolt holes should be slotted, with sliding keys to give uniform bearing pressure on slots under
transverse loads, as described above. In addition, each corner should have one large tapped hole for a
leveling/jacking screw that will allow bearing pad replacement, in situ.
Bolt shear stress from seismic acceleration
The maximum horizontal acceleration from a seismic event is expected to be much less than 1 m/s2; we use a
design value here of:
ahoriz 2m
s2
:=
Fhoriz Mtot ahoriz⋅:= Fhoriz 2.4 104
× N=
Bolt area required:
We calculate for all horizontal load taken on two corners only, since we will have sliding supports. We calculate for
austenitic stainless steel bolts:
Ssup_bolt Sf:= Ssup_bolt 137.895 MPa=
maximum shear stress:
Ss_sup_bolt 0.5Ssup_bolt:=
bolt area required, per corner
Asup_bolts
0.5Fhoriz
Ss_sup_bolt
:= Asup_bolts 1.74 cm2
=
assume 2 bolts per corner, for redundancy and symmetry about support web. with 2 bolts, the only critical
dimension to match between the holes in the support and the holes in the seismic frame are the distance
between the hole pairs (hole pattern rotation need not be matched). The sliding keys can be custom machined if
needed to compensate for mismatch.
length width changes from pressure:ESS_aus 193 GPa=
δLs
σ long Ls⋅
ESS_aus
:= δLs 0.28 mm=
δws
σhoop ws⋅
ESS_aus
:= δws 0.66 mm= in reality, the support itself will restrain a significant portion of this
deflection, since the saddle is welded to the vessel shell
thermal growth, 150C bakeout
αSS 16 106−
⋅ K1−
:= up to 100C
∆Tv 150K 20K−:=
ε th_SS αSS ∆Tv⋅:=
δv_t ε th_SS ws⋅:= δv_t 2.496 mm=
δv_l ε th_SS Ls⋅:= δv_l 2.122 mm=
bakeout will only be performed under vacuum condition.
Saddle support bending stress
Use an M24-2 bolt at each corner. Lubricate or PTFE coat (preferred)
djs_root 20.061 mm=djs_root4
πAjs:=
Ajs 3.161cm2
=Ajs
Fjs
0.9 Sy_316Ti⋅:=
Sy_316Ti 30000psi:=
Use 90% yield strength as allowable stress (non critical)
Fjs 1.323 104
× lbf=Fjs 5.884 104
× N=Fjs 0.5Mtot g⋅:=
Each jacking screw must be able to lift half the entire weight of the detector. We look for a low grade bolt that can
support this force
Jacking screw diameter
We choose only unfilled plastics, as most fillers are not radiopure (possible exception: bronze filled PTFE). PTFE
(unfilled), @500 psi , has little margin for stability, but any creep flow will act to equalize pressure over all 4
supports, resulting in a lower, stable pressure. Furthermore it is the only material that can withstand 150C,
although the temperature at the supports will be substantially less than 150C, due to the poor thermal
conductivity of SS. Cooling of supports should be performed in case of bakeout, regardless. Bronze-filled PTFE,
UHMWPE (non-oil-filled), nylon, or acetal may also be used; cooling of support pads during bakeout would be
mandatory.
Material for Bearing Pad
frrom Slideways bearing catalogue (similar to table 10-4 in J. Shigley, Mech. Engin. 3rd ed.)
Maximum allowable bearing pressures and temperatures (we may bake vessel at 150C with copper shielding inside)
Pbearing 425.162 psi=Pbearing 0.5Mtot g⋅
Abearing
:=
Bearing pressure is then (assuming a non-leveled condition where full weight is supported on two diagonal corners):
Abearing 200.724 cm2
=Abearing b12
4Asup_bolts−:=
Assume a full square contact patch under each corner;accounting for bolts and keys:
Bearing design
Support uses (2) M16-2.0 bolts on each corner, root diameter is 12mm
this is required minimum root diameterdsup_bolt 10.526 mm=dsup_bolt4
π0.5⋅ Asup_bolts:=
Iw
tw hw3
⋅
12:= Ifll
wfll tfll3
⋅
12:=
Iw 64 cm4
= Ifll 9.608 cm4
=Ir 1.201 cm
4=
dr 0.5tr c1−:= dw tr 0.5 hw⋅+( ) c1−:= dfll tr hw+ 0.5 tfll⋅+( ) c1−:=
dr 5.935− cm= dw 1.435− cm= dfll 3.565 cm=
Is_min Ir Ar dr2
⋅+
Iw Aw dw
2⋅+
+ Ifll Afll dfll
2⋅+
+:=
Is_min 973.459 cm4
=
Consider as a uniformly loaded beam, simply supported on each end
load per unit width (along the long dimension; transverse to vessel axis)
Mtot 1.2 104
× kg=ω
0.55Mtot g⋅
ws
:= ω 539.366N
cm=
Moment at center:
Msup_max
ω ws2
⋅
8:= Msup_max 9.709 10
3× N m⋅=
Maximum stress, tensile in flange under vessel
σsup_max
Msup_max 0.5⋅ hw
Is_min
:= σsup_max 39.893 MPa=
This is low enough to allow support only at corners; we do not need to support under the full width of the support
feet.
Cross section of saddle support is an I-beam, with a central "web" connecting two "flanges" We check bending
stress in support at bottom of vessel, where cross section height is a minimum.
Ro_pv Ri_pv tpv+:= Ro_pv 69 cm=Vessel axis height (axis above floor)
hv 80cm:=
flange and web thicknesses, widths:
tfll 2cm:= wfll b1:= wfll 14.411 cm= tw 1.5cm:= tr 1 cm=
I-beam web height, not including flanges:
hw hv Ro_pv tr+ tfll+( )−:= hw 8 cm=
I-beam Area Moment of Inertia:
Parallel axis theorem
sum moments of flanges and web about axis thru top surface , then divide by total area to find neutral axis
Ar b1 tr⋅:= Aw tw hw⋅:= Afll wfll tfll⋅:=
c1
Ar 0.5 tr⋅( )⋅ Aw tr 0.5 hw⋅+( )⋅+ Afll tr hw+ 0.5 tfll⋅+( )⋅+
Ar Aw+ Afll+:= c1 6.435 cm= down from top surface
Ir
b1 tr3
⋅
12:=
Sdiv1 1.379 108
× Pa=
Appendix 1-4 mandatory Supplemental Design Formulas
UG-32 does not give equations for a range of crown and knuckle radii; these are found in App 1-4
Lcr
rkn
10=
M1
43
Lcr
rkn
+
:= M 1.541=
Minimum shell thickness:
tmin
P Lcr⋅ M⋅
2S E⋅ 0.2P−:= tmin 11.871 mm= (3)
note: we will need full weld efficiency for the above thickness to be permissible, as per UG-32(b)
this formula is only valid if the following equation is true ( 1-4(a))
tmin
Lcr
8.729 103−
×=tmin
Lcr
0.002≥ 1=
Set head thickness:
th 12mm:=
Note: under EN_13455-3 rules for 316Ti, a thinner thickness of 10.25 mm is possible, due to a higher maximum
allowable strength at the knuckle. Below is an analysis from Sara Carcel
ANGEL Torispheric Head Design, using (2010 ASME PV Code Section VIII, div. 1, UG-32 Formed
heads and sections, Pressure on Concave Side, Appendix 1-4 rules eq 3
P 1.561 106
× Pa= E 1= S 2 104
× psi=
I.D.
Di 2Ri_pv:=
O.D.
Do Di 2t+:= Do 1.443 m=
Knuckle radius: Crown radius:
Lcr 1Di:= Lcr 1.36 m= rkn 0.1Di:= rkn 0.136 m=
Sy_316Ti 206.843 MPa=E 1= Sdiv1 20000psi:=
Torispherical heads, VIII, Div 2 DIN 28011 EN 13445-3 (316Ti, para DIN 28011
KORBBOGEN r=0,1L f 166.666667
D 1360 1360 Diámetro interior X 0.1
t 10 10 t 10
De 1380 1380 Y 0.00735294
L 1104 1360 Diámetro interior corona Z 2.13353891
ri 212.52 136 N 0.84954918
L/D 0.81176471 Ok 1 Entre 0,7 y 1, ver 4-49 β0,1 0.86799204
ri/D 0.15626471 Ok 0.1 Mayor de 0,06 β0,2 0.51421113
Li/t 110.4 Ok 136 Entre 20 y 2000 β 0.86799204
β 1.01880199 1.11024234 P 1.52
φ 0.49440713 0.85749293 eb 8.66383993
R 752.567792 697.850818 Si φ<β ey 10.2275849
C1 0.71313518 r/D>0,08 0.6742 es 6.21577196
C2 1.05371176 1.2 Thickness 10.2275849
Peth 64.2498476 E=117000 44.2387943
C3 206 Sy=206MPa 206
Py 3.37117586 1.57121205
G 19.0585868 28.1558395
Pck 6.73919067 3.14731728 G>1
lWne lshp:=
Fast vent reaction force, as calculated below
Ffv 3700N:= worst case is venting upward, at right angles to nozzle axis (we plan to use a straight through
valve, regardless, for which reaction force will not produce a bending moment and will simply
reduce longitudinal stress from pressure)Moments:
Mshp Fshp lshp⋅:= Mshp 113 N m⋅=
MWne Wne lWne⋅:= MWne 51.1 N m⋅=
Mfv Ffv Lne⋅:= Mfv 2146 N m⋅=
Total moment:
Mn Mfv Mshp+ MWne+:= Mn 2310 N m⋅=
Moment of Inertia, bending
In π Rn 0.5tn+( )3
⋅ tn⋅:= In 235.7 cm4
=
Stress, bending (longitudinal)
σn_l
Mn Rn tn+( )⋅
In
:= σn_l 50 MPa=
Stress, circumferential ( hoop)
Nozzle wall thickness required
Internal radius of finished opening
Rn 4.4cm:=
Thickness required for internal pressure:
trn
P Rn⋅
S E⋅ 0.6 P⋅−:= trn 0.501 mm=
We set nozzle thickness
tn 7mm:= we are limited by need to maintain CF bolt pattern which has typically a 4.0 inch OD pipe with room
for outside fillet weld
Don 2 Rn tn+( ):= Don 4.016 in=
Thickness required for external load
Nozzles on head may be subject to several possible non-pressure loads, simultaneously:
1. Reaction force from pressure relief, (fire) or fast depressure (auxiliary nozzle only)
2. Weight of attached components, including valves, expansion joints, copper or lead shielding plugs, high
voltage feedthrough.
The nozzles may all have nozzle extensions rigidly attached which create to possibility of high moments being
applied to the nozzles, not just shear loads. We consider the direction and location of center of gravity for these
loads:
ρPb 11.3gm
cm3
:=Lne 58cm:=
Forces and centers of gravity (l):
Fshp πRn2
Lne⋅ ρPb⋅ g⋅:= Fshp 391 N= lshp 0.5Lne:=(factor of 2 to account for flange weights)
Wne 2 2πRn Lne⋅ tn⋅ ρSS⋅( )⋅ g⋅:= Wne 176 N=
UG-36 (c) (3) Strength and Design of finished Openings:
Reinforcement is not required for the DN40 and DN75 flanged nozzles welded to the main cylindrical vessel as per
UG-36 below:
UG-37 Reinforcement Required for Openings in Shells and heads
OKPa_ne 93.795 bar=Pa_ne
4Bne
32Rn
tne
:=
Bne 13000psi:=Ane .02:=
From charts HA-1 and HA-2 above:
2Rn
tne
12.571=Lne
2Rn
6.591=
tne 7mm:=Lne 58 cm=
Nozzles on head are very short; no analysis needed. Nozzle extensions are longer:
External pressure:
σn_c
P Rn⋅
tn
:= σn_c 9.811 MPa=
Criterion for acceptable stress - use maximum shear stress theory:
Maximum shear stress (min. stress is in third direction, = zero on outside of nozzle):
τn
σn_l 0MPa−
2
2
:= τn 25 MPa= OK (J. Shigley, Mech.Eng. 3rd ed., eq. (2-9)
Compare with maximum shear stress from minimum thickness nozzle (pressure only, no applied moments)
σrn
P Rn⋅
trn
:= σrn 137 MPa=
τ rn
σrn 0MPa−
2
2
:= τ rn 68.5 MPa=
Additional Factor of Safety, over ASME factor of safety:
FSn
τ rn
τn
:= FSn 2.7= OK
<--no rapid fluctuations, condition met
<-- applicable to cyl. vessel,
condition met for DN40; DN75
nozzles<-- not applicable to cyl. vessel, but is
applicable to heads; condition not met
for DN100 nozzles, reinforcement needed
<--not applicable
<-- condition met
<-- condition met
In addition, there are no significant external loads on the radial nozzles of the vessel, only the weight of an HV
feedthrough at 45 deg angle; this is insignicant compared to the maximum loads and moments possible on the
head nozzles (which are similar in size and thickness). Proceeeding with the head nozzle reinforcement:
Reinforcement for the DN100 flanged nozzles welded to the torispheric heads is required and calculated according
to UG-37 :
Dp 0.158 m= (from UG-40 Limits of Reinforcement) fr4 1:= lege .71 te⋅:=
Area or reinforcement required:
Areq d tr⋅ F⋅ 2tn tr⋅ F⋅ 1 fr1−( )⋅+:= Areq 1.056 103
× mm2
=
Area available in shell:
A1a d E1 t⋅ F tr⋅−( )⋅ 2 tn⋅ E1 t⋅ F tr⋅−( )⋅ 1 fr1−( )⋅−:=
A1b 2 t tn+( )⋅ E1 t⋅ F tr⋅−( )⋅ 2 tn⋅ E1 t⋅ F tr⋅−( )⋅ 1 fr1−( )⋅−:=
A1 max A1a A1b,( ):= A1 0 mm2
=
Area available in nozzle projecting outwards
A2a 5 tn trn−( ) fr2⋅ t⋅:= A2a 389.914 mm2
=
A2b 2 tn trn−( )⋅ 2.5tn te+( )⋅ fr2⋅:= A2b 383.415 mm2
=
A2 min A2a A2b,( ):= A2 383.415 mm2
=
Area available in nozzle projecting inwards
nozzle and reinforcement are also 316Ti
trn 0.501 mm= d 2Rn:= d 8.8 cm= F 1.0:= fr 1.0:= fr1 1.0:= E1 1:= fr2 1:=
tr th:= tr 12 mm= ti 0mm:= h 0mm:= legi 0mm:= t 12mm:= legn 1.4tn:= legn 0.98 cm=
We will need a reinforcing pad, as the head is already minimum thickness and the nozzle is much thinner. Note: this
is for ASME; sec VIII, div. 1. European Codes allow thinner thickness for 316Ti.
te 12mm:= Dp 1.8d:=
A1 A2+ A3+ A41+ A42+ A43+ A5+ Areq≥ 1=
Torispheric Head, per DIN
A thinner head thickness of 10.58mm is calculated by S. Carcel to DIN formula; this is acceptable. It is not yet
clear whether or not reinforcement pads are needed.
Pressure Relief Capacity requirements
There are two possible conditions 1. regulator failure and 2 external fire
Lpv 2.1m:= length of vessel , inside average Ro_pv 0.69 m= outer radius
Pressure vessel outer area:
Apv 2πRo_pv2
2πRo_pv Lpv⋅+:= Apv 12.096 m2
=
From Anderson Greenwood Technical Seminar Manual, fire sizing is:
P1
MAWPpv
psi:=
F' .045:= A'Apv
ft2
:= A' 130.198=Aorif
F' A'⋅
P1
in2
⋅:=
P1 226.38=
KD 1:=Aorif 0.389 in
2= k 1.667:=
dorif4
πAorif⋅:= dorif 0.795 in=
A3a 5t ti⋅ fr2⋅:= A3a 0 mm2
=
A3b 5ti ti⋅ fr2⋅:= A3b 0 mm2
=
A3c 2 h⋅ ti⋅ fr2⋅:= A3c 0 mm2
=
A3 min A3a A3b, A3c,( ):= A3 0 mm2
=
Area available in weld, outward
A41 legn2
fr2⋅:= A41 96.04mm2
=
Area available in outer element weld
A42 lege2
fr4⋅:= A42 72.59mm2
=
Area available in weld, inward
A43 legi2
fr2⋅:= A43 0 mm2
=
Area available in reinforcement
A5 Dp d− 2tn−( ) te⋅ fr4⋅:= A5 676.8 mm2
=
Total Area available
A1 A2+ A3+ A41+ A42+ A43+ A5+ 1229 mm2
=
Area required:
Areq 1056 mm2
=
FT 3.694 103
× N=FT
W'k T⋅
k 1+( ) Ma⋅⋅
366lbf⋅:=
W' Whr
lb⋅:=
P2 0:=
Reactive force, from same ref. (pg. 49)
W 1.938 105
×lb
hr=W 24.419
kg
s=W
Ao Cg⋅ KD⋅ P'⋅ Ma⋅
T Zc⋅
lb
hr⋅:=
Zc .95:=
Ao
Avent
in2
:=Cg 377.641=Cg 520 k
2
k 1+
k 1+
k 1−
⋅⋅:=
Ma Ma_Xemol
gm⋅:=
R, ambientT 535:=Mass flow:P'
P
psi:=
Avent 4.383 in2
=Avent π 30mm( )2
⋅:=
However we will want to use the higher value which gives a fast vent , so as to safe Xe in case of leak
Tolerance analysis
Flange bolt holes:
First consideration is to realize that flanges absolutely must mate and bolt up without interference. This
means that tolerances must not be considered to add up in any statistical manner, all features must be
considered as being both at their limits of positional tolerance (oppositional) and in their maximum material
condition (MMC). That is, all holes and female features are as small as the tolerances allow, and all bolts and
male features are as large as the tolerances allow. Materials are all similar, so temperature ranges need not be
considered, but part deformations under load must be factored in.
Heads will be assembled to the vessel by first mounting them to an adjustable cradle support which
allows precise motions in all 6 degrees of freedom ( translations in x,y, and z, plus pitch, yaw and roll about the
center axes). This lift fixture is mounted on roller slides that move along the central vessel axis. The head is not
assembled to the vessel by hanging it loosely from a crane hook, though this, and other methods are acceptable
during construction.
A desirable, but not mandatory, design goal here is to assure that once the shear lip is assembled to the
vessel ID, the bolts will all insert without further translational alignment (rotation may still be needed). Thus, if the
vessel ID and mating shear lip are at MMC, and there is no remaining clearance between them, then the total bolt
hole tolerance is equal to the hole to bolt diametral clearance, at MMC. This must be shared between the head
and flange holes so the tolerance for each will be half the diametral tolerance. That is: for a 2 mm diametral
clearance, each bolt hole axis may be as much as 1.0 mm off its nominal position; the total will be no more than 1
mm which produces an effective aperture 2 mm smaller than the hole diameter, = 16 mm and bolts will still
assemble. In other words, the hole axis must be within a 1 mm radius (2 mm diameter) circle, thus the true
position tolerance for the bolt holes is (a cylinder of) 2 mm dia. this is illustrated below:
dcl_mmc 18mm:=
db 16 mm= note root diameter is less, but threaded portion is 25 mm long and must pass through both
holes simultaneously
We need to account for vessel flange deflection under load which will distort the hole pattern. Maximum deflection,
in the vertical direction is:
δfl 0.1mm:=
This is from an ANSYS workbench model with a 60000N load applied to each vessel internal flange,
no head present. Assume head flange is undistorted
( )
Maximum offset of bolt holes for flanges at LMC, bolts and holes at MMC
∆rcl 1.5 mm=∆rcl Ri_pv ∆Ri_pv+( ) Rsl ∆Rsl−( )−:=
Maximum offset of shear lip and vessel ID axes (both at LMC)
If false, additional shift of head relative to vessel may be necessary, even though shear lip assembles to vessel
ID. If true, we can proceed to check for the case of shear lip and vessel ID at LMC, below:
tbh_acc tbh_max≤ 0=Check:
Will head and bolts "auto-assemble" (assemble without further translational alignment) for shear lip and vessel ID
at MMC?
tbh_acc 2mm:=
Using this value might require a very high alignment precision to find the proper bolt alignment so we use a
slightly smaller value
tbh_acc_max 2.3 mm=tbh_acc_max tbh 2 rcl δfl−( )+:=
The radial clearance between shear lip and vessel ID (both at MMC) is represents an additional tolerance that we
can add to the accuracy tolerance, because we can use it to shift the patterns to match. Since tolerances are
specified on a diameter basis, we add 2x the radial offset (minus 2x the deflection):
Head will assemble to flange with full detector mass loading (safety factor>2)rcl 2δfl> 1=Check:
rcl 0.5 mm=rcl Ri_pv ∆Ri_pv−( ) Rsl ∆Rsl+( )−:=
Nominal radial clearance between shear lip and vessel, both at MMC:
∆Rsl 0.25mm:=Rsl 679mm:=(+/-)
(+/-)∆Ri_pv 0.25mm:=Ri_pv 680 mm=
tbh tbh_max< 1=tbh 1.5mm:=
Set:
This is total maximum positional tolerance diameter for each flange hole, assuming the nominal hole positions of
head and vessel flanges are in alignment.
There are two ways to specify bolt hole positional tolerance, either with respect to themselves as a pattern
(the pattern otherwise unconstrained) or each hole individually, with respect to the specified datums. The former is
a precisional tolerance, the latter an accuracy tolerance, (which is more difficult to achieve).
For the case where there is no remaining clearance between the shear lip and the vessel ID, when both
are at MMC, the requirements for accuracy and precision are the same, and tbh_max is the maximum allowable
positional tolerance with respect to the datums A/B,C/D, and E/F; that is we have only an absolute accuracy
requirement for bolt hole positional tolerance.
Any radial clearance between the shear lip and the vessel ID, with both at MMC, allows the two hole
patterns to shift, as an ensemble, with respect to each other. This allows a larger maximum allowable positional
tolerance of the holes with respect to the datums A/B,C/D, and E/F (accuracy requirement), but still requires that
the hole pattern still be toleranced to tbh_max or smaller, with respect to itself (a local precision or repeatability
requirement). This is accomplished with two tolerance blocks on the drawing. The drawback is that the shear lip
and vessel id may assemble, but in a shifted condition, such that bolts will not assemble. Additional alignment will
then be needed. The cure for this is to tighten the tolerance (from tbh_max ) on the hole pattern in reference to itself,
by the maximum shift that can occur when the shear lip and vessel ID are in LMC condition. Given that these are
large features, their tolerance will necessarily be large.
tbh_max 1.9 mm=
tbh_max dcl_mmc db δfl+( )−:=
(+/-)∆θbh_acc 0.06 deg=∆θbh_acc.71
2
tbh_acc
Ri_pv
:=
(+/-)∆rbh_acc 0.71 mm=∆rbh_acc.71
2tbh_acc:=
(accuracy tolerance)With respect to datums A/B,C/D,E/F
(+/-)∆θbh 0.045 deg=∆θbh.71
2
tbh
Ri_pv
:=
(+/-)∆rbh 0.532 mm=∆rbh.71
2tbh:=
With respect to each other (precisional tolerance)
Equivalent maximum radial, angular misalignments of bolt holes (given here as +/- values) for all parts at MMC.
These describe square tolerance zones inscribed within the circular tolerance zones
We conclude that we may need to further translate and rotate the head relative to the vessel in order to align the
bolt holes, even though the shear lip assembles. Since no "autoassembly is possible, we can loosen the
accuracy requirement conditionally by specifying the true position tolerance for circular datum C or D at MMC
condition; this allows the final bolt hole accuracy tolerance to increase by the amount datum C or D are from
MMC.
The head must be retracted for this operation as the actual clearance between the shear lip and the
vessel ID will not be known. Furthermore, the adjustment of the struts is not performed simultaneously, and large
intermediate translations or rotations of the headmay take place prior to achieving the final small alignment. This
would cause an interference of the shear lip with the ID, with possible damage. Internal components may require
further retraction of head prior to alignment. LBNL Engineering Note 10182B, D. Shuman, provides a general
method and MathCAD worksheet for determining the needed strut adjustments to align a component. The 6 strut
head alignment and assembly fixture designed for the heads has Cartesian motions which are largely uncoupled
and should be simple enough to adjust intuitively without needing this methodology.
∆rcl tbh+ tbh_max≤ 0=
check if bolts will assemble with vessel ID and shear lip assembled at LMC (fully misaligned), without further
translation alignment:
∆rcl tbh+ 3 mm=